Question 1170424: 𝐏𝐫𝐨𝐛𝐥𝐞𝐦 𝟏
(a) Teddy J is a manufacturer of dish washing liquid . If his monthly demand function for 750ml size is q = 4000 − 250p and his total cost function is C(q) = 500 + 0.2q.
(i) Derive an expression, R(q) for Teddy J′s total revenue curve.
(ii) Derive an expression, Π(q) for Teddy J′s profit function.
(iii) Determine whether Teddy J′s profit is increasing or decreasing when he produces 5 hundred, 750ml bottles of dish washing liquid.
(iv) How many 750ml bottles of dish washing liquid should Teddy J produce per month if he wishes to maximize his profits.
(b) A firm has an average cost function
125 q2
A(q) = + − 4. q 16
where q is the firm′s output.
(i) Determine the level of output for average costs are minimum.
(ii) Hence determine the range of values for which average costs are decreasing.
(iii) What part of the decreasing range is practically feasible?
(iv) Write an equation for the total cost function.
(v) Hence calculate the level of output for which total costs are minimum.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down each part of the problem.
**(a) Teddy J's Dish Washing Liquid**
Given:
* Demand function: $q = 4000 - 250p$
* Total cost function: $C(q) = 500 + 0.2q$
**(i) Total Revenue (R(q))**
* First, solve the demand function for $p$:
* $250p = 4000 - q$
* $p = \frac{4000 - q}{250} = 16 - \frac{q}{250}$
* Total revenue is given by $R(q) = pq$:
* $R(q) = \left(16 - \frac{q}{250}\right)q$
* $R(q) = 16q - \frac{q^2}{250}$
**(ii) Profit Function (Π(q))**
* Profit is total revenue minus total cost:
* $\Pi(q) = R(q) - C(q)$
* $\Pi(q) = \left(16q - \frac{q^2}{250}\right) - (500 + 0.2q)$
* $\Pi(q) = 16q - \frac{q^2}{250} - 500 - 0.2q$
* $\Pi(q) = -\frac{q^2}{250} + 15.8q - 500$
**(iii) Profit at q = 500**
* To determine if profit is increasing or decreasing, we need to find the derivative of the profit function and evaluate it at $q = 500$:
* $\Pi'(q) = -\frac{2q}{250} + 15.8 = -\frac{q}{125} + 15.8$
* $\Pi'(500) = -\frac{500}{125} + 15.8 = -4 + 15.8 = 11.8$
* Since $\Pi'(500) > 0$, the profit is increasing when Teddy J produces 500 bottles.
**(iv) Profit Maximization**
* To maximize profit, we set the derivative of the profit function to zero:
* $\Pi'(q) = -\frac{q}{125} + 15.8 = 0$
* $\frac{q}{125} = 15.8$
* $q = 15.8 \cdot 125 = 1975$
* Teddy J should produce 1975 bottles to maximize profits.
**(b) Firm's Average Cost Function**
Given:
* Average cost function: $A(q) = \frac{125}{q} + \frac{q}{16} - 4$
**(i) Output for Minimum Average Costs**
* To find the minimum average cost, we take the derivative of $A(q)$ and set it to zero:
* $A'(q) = -\frac{125}{q^2} + \frac{1}{16} = 0$
* $\frac{125}{q^2} = \frac{1}{16}$
* $q^2 = 125 \cdot 16 = 2000$
* $q = \sqrt{2000} = 20\sqrt{5} \approx 44.72$
* The level of output for minimum average costs is approximately 44.72.
**(ii) Range for Decreasing Average Costs**
* Average costs are decreasing when $A'(q) < 0$:
* $-\frac{125}{q^2} + \frac{1}{16} < 0$
* $\frac{1}{16} < \frac{125}{q^2}$
* $q^2 < 125 \cdot 16 = 2000$
* $q < \sqrt{2000} = 20\sqrt{5} \approx 44.72$
* Average costs are decreasing when $0 < q < 20\sqrt{5}$.
**(iii) Practically Feasible Decreasing Range**
* Output must be a positive value.
* The practically feasible range for decreasing average costs is $0 < q < 20\sqrt{5}$.
**(iv) Total Cost Function (C(q))**
* Total cost is average cost multiplied by output:
* $C(q) = q \cdot A(q)$
* $C(q) = q \left(\frac{125}{q} + \frac{q}{16} - 4\right)$
* $C(q) = 125 + \frac{q^2}{16} - 4q$
**(v) Output for Minimum Total Costs**
* To find the minimum total cost, we take the derivative of $C(q)$ and set it to zero:
* $C'(q) = \frac{2q}{16} - 4 = \frac{q}{8} - 4 = 0$
* $\frac{q}{8} = 4$
* $q = 32$
* The level of output for minimum total costs is 32.
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